# SOLUTION: I REALLY REALLY NEED HELP A.S.A.P. What is the value of N in the equation: log{{{8}}}(n-3) + log{{{8}}}(n+4) = 1. I REALLY REALLY appreciate the help because I really need it. Sorr

Algebra ->  Algebra  -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I REALLY REALLY NEED HELP A.S.A.P. What is the value of N in the equation: log{{{8}}}(n-3) + log{{{8}}}(n+4) = 1. I REALLY REALLY appreciate the help because I really need it. Sorr      Log On

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 Question 20724: I REALLY REALLY NEED HELP A.S.A.P. What is the value of N in the equation: log(n-3) + log(n+4) = 1. I REALLY REALLY appreciate the help because I really need it. Sorry for the short notice.Found 2 solutions by venugopalramana, kapilsinghi:Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!i think you mean logs to base 8 ...assuming that and using the formula log x to base y = log x /log y...where all logs are taken to any common base. so we have log(n-3)/log 8 + log(n+4)/log 8 =1....multipling through out with log 8,we get log (n-3)+log (n+4)=log 8 log(n-3)(n+4)=log 8 hence (n-3)(n+4)=8 n^2-3n+4n-12-8=0 n^2+n-20=0 n^2+5n-4n-20=0 n(n+5)-4(n+5)=0 (n+5)(n-4)=0 hence n=-5 or n=4 Answer by kapilsinghi(68)   (Show Source): You can put this solution on YOUR website!log(n-3) + log(n+4) = 1 log(n-3) + log(n+4) = log(8) log(n-3)(n+4) = log(8) taking antilog on both sides n^2 + n - 12 = 8 n^2 + n - 20 = 0 (n +5)(n-4)=0 n = -5 or n = 4 by substitution n=4 seems correct kapilsinghi123@gmail.com